Skip to main content

const overload functions taking char pointers

Always have two overloaded versions of functions that take
char * and const char * parameters. Declare (but don't define if not needed)
a function that takes const char* as a parameter when you have defined
a function that accepts a non-const char* as a parameter.

#include <iostream>
#include <iomanip>

static void foo (char *s) {
std::cout << "non-const " << std::hex << static_cast <void *>(s) << std::endl;
}
static void foo (char const *s) {
std::cout << "const " << std::hex << static_cast <void const *>(s) << std::endl;
}

int main (void)
{
char * c1 = "Literal String 1";
char const * c2 = "Literal String 1";
foo (c1);
foo (c2);
foo ("Literal String 1");
//*c1 = 'l'; // This will cause a seg-fault on Linux.
std::cout << c2 << std::endl;

return 0;
}

Because of default conversion rule from string literal to char *,
the call to foo using in-place literal goes completely undetected
through the eyes of compiler's type system.

Interestingly enough, the addresses of all the identical string literals
is the same, irrespective of whether it is assigned to const or non-const.
Internally though, they are stored on the const DATA page and modifying
them causes a seg-fault.

Comments

Steven said…
Hi there,


Could you explain why the *c1 = 'l' segfaults under Linux? To me it looks like modifying a proper (non const) string?


Thanks!
Sumant said…
"Literal String 1" is the same for all the char pointers, const or non-const. The compiler detects uniqueness of the initializer strings and creates exactly one copy of it in the DATA pages of the process, which are read only. So const and non-const both point to the same const readonly string and therefore an attempt to modify it seg-faults. I think irrespective to what a string literal is assigned, a string literal will always be in the DATA page readonly.
Suman Kar said…
What you ascribe to the uniqueness of the literals is only a part answer.It would crash even if there was only one non const char pointer to a literal and you attempted to modify the literal's value. The Standard defines it to be an Undefined behaviour and hence *anything* might happen. Some compilers have the option of making literals writable (Gcc is a case in point).

The rational behind such a decision can be attributed to a number of reasons. One example is, as you've said, the literals being in the DATA segment -- modifying the same is tantamount to creating a self-modifying program.
Sorry, I did not understand how the fact that a string literal converts to char* infers that one needs two overloads for char* and char const*?
Sumant said…
@maxim
When you declare a function taking a non-const char pointer, you are also declaring your intention to potentially change the input parameter. If you really are sure that you would never change the input parameter string use const char pointer instead. Now assuming that you do intend to change the input parameter string, you declared it as a non-const char pointer and someone out there passed in a string literal to it. That is a problem. Your function will try to change the string literal which is undefined by standard. To avoid this from happening you can take some help from the compiler. If you just declare a function of the same name and rest of the signature, taking const char pointer as input parameter, compiler will pick that up when someone out there uses a string literal to call it. const char pointer is a better match than non-const char pointer for a string literal. But because it is undefined (remember you just declared it) program won't link.
Amber said…
Of course with modern compilers you **will** get a warning:

test.cpp|16 col 17| warning: deprecated conversion from string constant to ‘char*’ [-Wwrite-strings]

so this advice is a bit obsolete
Sumant said…
Some people still need old compilers. I do.

Popular Content

Multi-dimensional arrays in C++11

What new can be said about multi-dimensional arrays in C++? As it turns out, quite a bit! With the advent of C++11, we get new standard library class std::array. We also get new language features, such as template aliases and variadic templates. So I'll talk about interesting ways in which they come together. It all started with a simple question of how to define a multi-dimensional std::array. It is a great example of deceptively simple things. Are the following the two arrays identical except that one is native and the other one is std::array? int native[3][4]; std::array<std::array<int, 3>, 4> arr; No! They are not. In fact, arr is more like an int[4][3]. Note the difference in the array subscripts. The native array is an array of 3 elements where every element is itself an array of 4 integers. 3 rows and 4 columns. If you want a std::array with the same layout, what you really need is: std::array<std::array<int, 4>, 3> arr; That's quite annoying for

Unit Testing C++ Templates and Mock Injection Using Traits

Unit testing your template code comes up from time to time. (You test your templates, right?) Some templates are easy to test. No others. Sometimes it's not clear how to about injecting mock code into the template code that's under test. I've seen several reasons why code injection becomes challenging. Here I've outlined some examples below with roughly increasing code injection difficulty. Template accepts a type argument and an object of the same type by reference in constructor Template accepts a type argument. Makes a copy of the constructor argument or simply does not take one Template accepts a type argument and instantiates multiple interrelated templates without virtual functions Lets start with the easy ones. Template accepts a type argument and an object of the same type by reference in constructor This one appears straight-forward because the unit test simply instantiates the template under test with a mock type. Some assertion might be tested in

Want speed? Use constexpr meta-programming!

It's official: C++11 has two meta-programming languages embedded in it! One is based on templates and other one using constexpr . Templates have been extensively used for meta-programming in C++03. C++11 now gives you one more option of writing compile-time meta-programs using constexpr . The capabilities differ, however. The meta-programming language that uses templates was discovered accidently and since then countless techniques have been developed. It is a pure functional language which allows you to manipulate compile-time integral literals and types but not floating point literals. Most people find the syntax of template meta-programming quite abominable because meta-functions must be implemented as structures and nested typedefs. Compile-time performance is also a pain point for this language feature. The generalized constant expressions (constexpr for short) feature allows C++11 compiler to peek into the implementation of a function (even classes) and perform optimization